UVA 10534 Wavio Sequence (双向LIS)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475

思路：以a[i]为Wavio Sequence的最大值，（根据定义）计算从0到i和从i到n的lis长度的较小者，为以a[i]为最大值的Wavio Sequence的半长度，最后取所有半长度的最大值*2-1

完整代码：

/*0.102s*/

#include<bits/stdc++.h>
using namespace std;
const int mx = 10005;

int a[mx], d[2][mx], lis[mx];

int main()
{
	int N, len, i, j, ans;
	while (~scanf("%d", &N))
	{
		len = 0;
		for (i = 1; i <= N; ++i)
		{
			scanf("%d", &a[i]);
			d[0][i] = j = lower_bound(lis + 1, lis + len + 1, a[i]) - lis;
			len = max(len, j);
			lis[j] = a[i];
		}
		len = 0;
		for (i = N; i; --i)
		{
			d[1][i] = j = lower_bound(lis + 1, lis + len + 1, a[i]) - lis;
			len = max(len, j);
			lis[j] = a[i];
		}
		ans = 0;
		for (i = 1; i <= N; ++i)
			ans = max(ans, (min(d[0][i], d[1][i]) << 1) - 1);
		printf("%d\n", ans);
	}
	return 0;
}